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A prism having refractive index $\sqrt{2}$ and refracting angle $30^{\circ}$ has one of the
refracting surfaces silvered. The beam of light incident on the other refracting
surface will retrace its path, if angle of incidence is $\left[\sin \frac{\pi}{6}=0 \cdot 5\right]$
Options:
refracting surfaces silvered. The beam of light incident on the other refracting
surface will retrace its path, if angle of incidence is $\left[\sin \frac{\pi}{6}=0 \cdot 5\right]$
Solution:
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Verified Answer
The correct answer is:
$\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
(D)
$A=r_{1}+r_{2}$ but $r_{2}=0$ since is incident normally on the second surface.
$\therefore \mathrm{r}_{1}=\mathrm{A}=30^{\circ}$
$n=\frac{\sin i}{\sin r}$
$\therefore \sqrt{2}=\frac{\sin \mathrm{i}}{\sin 30^{\circ}}$
$\therefore \sin \mathrm{i}=\sqrt{2} \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}$
$\therefore \mathrm{i}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
$A=r_{1}+r_{2}$ but $r_{2}=0$ since is incident normally on the second surface.
$\therefore \mathrm{r}_{1}=\mathrm{A}=30^{\circ}$
$n=\frac{\sin i}{\sin r}$
$\therefore \sqrt{2}=\frac{\sin \mathrm{i}}{\sin 30^{\circ}}$
$\therefore \sin \mathrm{i}=\sqrt{2} \sin 30^{\circ}=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}}$
$\therefore \mathrm{i}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$
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